A LEAD COMPENSATOR CHARACTERISTICS USING BODE DIAGRAM FOR MAXIMUM OF 50 DEG PHASE ANGLE THIS PPT IS SO USEFUL FOR THE ENGINEERING STUDENTS FOR CONTROL SYSTEMS STUDENTS AND THIS PPT ALSO CONTAINS A MATLAB CODING FOR THE LEAD COMPENSATOR AND THE RESULTS ARE ALSO PLOTTED IN THAT PPT AND THE PROBLEM CAN ALSO BE SOLVED BY USING THE DATA IN PPT AND IT IS SO USEFUL PPT

1. Raghu Engineering College
Department of Electrical & Electronics Engineering(Autonomous)
Accredited by NBA & NAAC with ‘A Grade, Permanently Affiliated JNTU Kakinada
Dakamarri (v), Bheemunipatnam Mandal, Visakhapatnam, Andhra Pradesh 531162
A CASE STUDY ON
A LEAD COMPENSATOR CHARACTERISTICS USING BODE DIAGRAM
FOR MAXIMUM OF 50 DEG PHASE ANGLE
UNDER THE GUIDANCE
Dr.B.SIVA RAMA RAO
ASSOCIATE PROFESSOR
BY
P. ESWAR SAI 18981A0237
P.MOHAN 18981A0238
P.SUPRIYA 18981A0239
P.NITHESH KUMAR 18981A0240
P.SANJAY KUMAR 18981A0241
BACHELOR OF TECHNOLOGY
IN
ELECTRICAL AND ELECTRONICS ENGINEERING

2. CONTENTS
1.Introduction
2.Phase Lead
4.Calculations
5.Design Approach
6.Design specific to Phase lead Compensator
7.Examples

3. INTRODUCTION
• A feedback control system that provides an optimum Performance
without any necessary adjustment is rare
• In building a control system , we know that proper Modification of
the plant dynamics may be simple way to meet the performance
specifications.
• This however, may not be possible in many practical situations
because the plant may be fixed and not modifiable.
• Then we must adjust parameters other than those in the fixed plant.

4. Compensation schemes commonly used for feed back
Control systems are :
-Series compensation
-parallel compensation

5. • The choice between series compensation and parallel
compensation
depends on
 The nature of the signals
 The power levels at various points
 Available components
 The designer’s experience
 Economic considerations and so on.

6. PHASE LEAD COMPENSATOR :-
 The magnitude of the compensator continuously grows with
increasing frequency.
• The feature is undesirable because it amplifies high frequency
noise that is typically present in an real system.
• A typical lead compensator has the following transfer function.
𝐶 𝑠 = 𝐾
𝜏𝑠+1
𝛼𝜏𝑠+1
where , 𝛼<1
•
1
𝛼
is the ratio between the pole zero break point frequencies.
• Magnitude of the lead compensator is 𝐾
1+𝜔2 𝜏2
1+𝛼2 𝜔2 𝜏2

7. • The phase contributed by the lead compensator is given by
𝜑 = tan−1 𝜔𝜏 − tan−1 𝛼𝜔𝜏
• Thus a significant amount of phase is still provided with much less amplitude at
high frequencies.
• The frequencies response of a typical lead compensator magnitude varies from
20 log10 𝐾 to 20 log10
𝐾
𝛼
and maximum phase is always less than 900
.
• The frequency can be shown where the phase is maximum
𝜔 𝑚𝑎𝑥 =
1
𝜏 𝛼
• The maximum phase corresponds to sin∅ 𝑚𝑎𝑥 = (
1−sin (∅ 𝑚𝑎𝑥)
1+sin(∅ 𝑚𝑎𝑥)
)
• The magnitude of C(s) at 𝜔 𝑀𝐴𝑋 is
𝐾
𝛼

8. Analyze the stability of the System with transfer function using
Bode Diagram
Given transfer function:
𝟒
𝑺(𝑺+𝟏) 𝟐
Simplify the given transfer function
Transfer function =
4
𝒔 𝟑+𝟐𝒔 𝟐+𝒔
Use the MATLAB codding to find out the stability of a given transfer function
by obtaining the bode graph.

9. •Then run the codding, this is the bode graph of the given system.

10. •From the above bode graph of the given transfer function is a unstable system because of
the phase margin and gain margin are the negative values.
•Gain margin = -6.02 dB
•Phase margin = -18.1 degree
•Natural frequency(Wn) = 1.38 rad/sec
•The gain margin and phase margin are negative value, then the system becomes
unstable.
•So by adding the some phase to the system then the phase margin of the system becomes
positive values. Then system becomes stable.
•Use the lead compensator for adding phase margin to the system.

11. 2.Design of lead compensator :-
 This is the lead compensator network. It consists of resistors and capacitor.
 The capacitor(c) is connected parallel to the R1.
 The output is taken across the R2.
 Before going to design the lead compensator we have to know the ‘α’ value.
 If we add the ‘500’ to the system then phase margin become positive.
 If we add more phase angle to the system then the system performance will
decrease. Else we add less phase angle to the system then the system stability
will decrease
 We already determine the formula for knowing the ‘α’ value.

12. α=
𝟏−𝒔𝒊𝒏∅
𝟏+𝒔𝒊𝒏∅
here ∅ is the how much have to added to the system.
Then ∅ = 500
α=
𝟏−𝒔𝒊𝒏(𝟓𝟎)
𝟏+𝒔𝒊𝒏(𝟓𝟎)
=
𝟏−𝟎.𝟕𝟔𝟔
𝟏+𝟎.𝟕𝟔𝟔
= 0.132
α=
R2
R1+
R2
-----------------(1)
Assuming the R2 value = 250ohm, substitute value in equation (1)
0.132 = 250
R1+25
0
0.132(R1+250 ) = 25
R1 =
250
−(
250
∗
0
.
132
)
0.132
R1 = 1643.93 ohm

13. R1 = 1643.93 ohm, R2 = 250 ohm
for determine the capacitor value
T = C * R1
C =
T
R1
----------(2)
We know the relation for determine the T value
Wn =
1
T√∝
----------(3)
Wn = 1.38 rad/sec
Substitute value in equation (3)
T=1Wn √∝
T=11.38 √0.198
T=1.99
T=1.99

14. Then substitute this value in equation (2)
C = T
R1
C =
1
.
99
1643.93
C = 1.2105 mF
After knowing the values of lead compensation then design the circuit. Use the
MATLAB codding for find out the bode graph of the lead compensator with the
above values.

15. Bode graph to the lead compensator.
Above graph is for compensator , here the phase margin is 50 degree angle. So,
add this compensator to our system. Then system becomes stable.

16. Compensated system:
 connect the compensator to the system in cascade connection.
 Phase margin of the system without compensator = -18.1 degree
 Compensator maximum phase angle = 50.1 degree
 Values of phase margin and gain margin when compensator added to the system by
using MATLAB. Shown in the below fig. is codding for compensated system in MATLAB.

17. Corresponding Bode Graph
By the bode graph
Gain margin = 0.23 dB
Phase margin = 0.711 degree
The gain margin and phase margin are positive values. Then the system become
stable.